Let $\sum _{k=0}^{\infty }c{r}^k$and$\sum _{k=0}^{\infty }b{v}^k$be two convergent geometric series. Prove that $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ converges. If neither c nor b is 0, could the series be $\left(\sum _{k=0}^{\infty }c{r}^k\right)\left(\sum _{k=0}^{\infty }b{v}^k\right)$?

$\sum _{k=0}^{\infty }c{r}^k$and$\sum _{k=0}^{\infty }b{v}^k$are two convergent geometric series.

When we expand the series $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$we get,

$$\begin{gathered} \sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)=cb+cbrv+..... \\ =cb\left(1+\left(rv\right)+{\left(rv\right)}^2+.....\right) \\ =cb\left(\sum _{k=0}^{\infty }{\left(rv\right)}^k\right) \end{gathered}$$

As, the above expression is in geometric progression. Therefore the expression $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$is also geometric.

Now as we know $\sum _{k=0}^{\infty }\left(c{r}^k\right)$converges which means $\left|r\right|<1$and also $\sum _{k=0}^{\infty }\left(b{v}^k\right)$converges which means $\left|v\right|<1$.

From this, it can be concluded that $\left|rv\right|<1$.

Now, look at the expression from previous step we get $cb\sum _{k=0}^{\infty }{\left(rv\right)}^k$.

And we found that $\left|rv\right|<1$.

So the expression $cb\sum _{k=0}^{\infty }{\left(rv\right)}^k$or $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$converges.

$$\begin{gathered} \sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)=cb\left(1+rv+{\left(rv\right)}^2+....\right) \\ \sum _{k=0}^{\infty }\left(c{r}^k\right).\sum _{k=0}^{\infty }\left(b{v}^k\right)=c\left(1+r+r^2+....\right)b\left(1+v+v^2+....\right) \end{gathered}$$

It is clear from above that the right hand side of the above two equations is not equal therefore the left hand sides also cannot be equal as well

Therefore, $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)\ne \sum _{k=0}^{\infty }\left(c{r}^k\right).\sum _{k=0}^{\infty }\left(b{v}^k\right)$.